Answer:
Explanation:
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In this case, since we have a problem about volume-pressure relationship, the idea here is to use the Boyle's law to calculate the final volume as shown below:
Then, we plug in the initial and final pressures and the initial volume to obtain:
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<u>Answer:</u> The percent yield of the water is 31.98 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of methane = 6.58 g
Molar mass of methane = 16 g/mol
Putting values in equation 1, we get:
Given mass of oxygen gas = 14.4 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the combustion of methane is:
By Stoichiometry of the reaction:
2 moles of oxygen gas reacts with 1 mole of methane
So, 0.45 moles of oxygen gas will react with = of methane
As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction
2 moles of oxygen gas produces 2 moles of water
So, 0.45 moles of oxygen gas will produce = moles of water
Molar mass of water = 18 g/mol
Moles of water = 0.45 moles
Putting values in equation 1, we get:
Experimental yield of water = 2.59 g
Theoretical yield of water = 8.1 g
Putting values in above equation, we get:
Hence, the percent yield of the water is 31.98 %